0 0 votes The surface area of the portion of the paraboloid $z=x^{2}+y^{2}$ that lies between the planes $z=0$ and $z=\frac{1}{4}$ is $\frac{\pi}{6}(2 \sqrt{2}-1)$ $\frac{\pi}{2}(2 \sqrt{2}-1)$ $\pi(2 \sqrt{2}-1)$ $\frac{\pi}{3}(2 \sqrt{2}-1)$ Functions of Two Variables gatexe-2023 functions-of-two-variables vector-calculus + – admin 8.4k points answer 0 reply
